Problem: $\sum\limits_{k=1}^{625 }{{(1-7k)}}=$
What is the question asking for? The question is asking for the sum of the values of $1-7k $ from $k = 1$ to $k = 625 $ : $(1-7 \cdot 1 ) + (1-7 \cdot 2) +... + (1-7\cdot {625} )$ The series is arithmetic because the formula $1-7k $ is a linear function of $k$. Formula for arithmetic series The sum $S_n$ of a finite arithmetic series is $S_n = \dfrac {\left(a_1 + a_n \right)}{2} \cdot n$ where $a_1$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. What do we need to use the formula? The number of terms $(n = {625})$ is the upper limit of the sigma notation. We need to find $a_1$ (the first term) and $a_{625}$ (the last term). Step 1: Find $a_1$ and $a_{625}$ (the first and the last term) $a_1 = 1-7(1) = {-6}$ $a_{625} =1 -7(625) = {-4374}$ Step 2: Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac {\left(a_1 + a_n \right)}{2} \cdot n \\\\ S_{{625}}&= \dfrac {\left({-6} + ({-4374}) \right)}{2} \cdot {625} \\\\ S_{{625}} &= -2190 \left(625\right) \\\\ S_{{625}} &= -1{,}368{,}750\end{aligned}$ The answer $ - 1{,}368{,}750 $